# Exercise 5.3Z: Realization of a PN Sequence

Two PN generator realizations

The diagram shows two possible generators for generating PN sequences in unipolar representation:   $u_ν ∈ \{0, 1\}$.

• The upper generator with the coefficients
$$g_0 = 1 \hspace{0.05cm}, \hspace{0.2cm}g_1 = 0 \hspace{0.05cm}, \hspace{0.2cm}g_2 = 1 \hspace{0.05cm}, \hspace{0.2cm}g_3 = 1 \hspace{0.05cm}$$
is denoted by the octal identifier   $(g_3,\ g_2,\ g_1,\ g_0)_{\rm octal} = (15)$.
• Accordingly,  the octal identifier of the second PN generator is  $(17)$.
• One speaks of an M-sequence if for the period length of the sequence   $〈u_ν〉$  holds:
$$P = 2^G – 1.$$
Here,  $G$  denotes the degree of the shift register,  which is equal to the number of memory cells.

Notes:

### Questions

1

What is the degree  $G$  of the two PN generators considered here?

 $G \ = \$

2

Give the period length  $P$  of the PN generator with the octal identifier  $(15)$.

 $P\ = \$

3

Which of the following statements are true for each M-sequence?

 The number of  "zeros"  and  "ones"  is the same. In each period there is one more  "ones"  than  "zeros". The maximum number of consecutive  "ones"  is  $G$. The sequence  $1 0 1 0 1 0$ ...   is not possible.

4

Specify the period length  $P$  of the PN generator with the octal identifier $(17)$.

 $P\ = \$

5

Which PN generator produces an M-sequence?

 The generator with the octal identifier  $(15)$. The generator with the octal identifier  $(17)$.

### Solution

#### Solution

(1)  The degree  $\underline{G = 3}$  is equal to the number of memory cells of the shift register.

(2)  From the given sequence the period length  $\underline{P = 7}$  can be read.  Because of  $P = 2^G –1$  it is an M-sequence.

(3)  Solutions 2, 3 and 4  are correct:

• The maximum number of consecutive  "ones"  is  $G$  (whenever there is a  "one"  in all  $G$  memory cells).
• On the other hand,  it is not possible that all memory cells are filled with zeros  (otherwise only zeros would be generated).
• Therefore,  there is always one more  "ones"  than zeros.
• The period length of the sequence  "$1 0 1 0 1 0$ ..."   is  $P = 2$.  For an M-sequence  $P = 2^G –1$.  For no value of  $G$:    $P = 2$  is possible.

(4)  If all memory cells are occupied with ones,  the generator with the octal identifier  $(17)$  returns a  $1$  again:

$$u_{\nu} \big [ u_{\nu-1} + u_{\nu-2} + u_{\nu-3} \big ] \,\,{\rm mod} \,\,2 =1 \hspace{0.05cm}.$$
• Since this does not change the memory allocation,  all further binary values generated will also be  $1$  each   ⇒   $\underline{P = 1}$.

• One speaks of an M-sequence only if  $P = 2^G –1$  holds.